This lecture is an example of how a system of linear equations can be used to derive formulas of special theory of relativity. Albert Einstein has derived these formulas in his "Electrodynamics" in a more physical, more intuitive way. This lecture is pure mathematics and, as such, causes much less problems in understanding.

1. Assume we have two systems of coordinates, one stationary with coordinates {X,T} (assuming for simplicity all the movements will occur in one space dimension along X-axis and one time dimension) and another with coordinates {x,t} moving along the X-axis with constant speed V.

2. Assume that at time T=0 systems coincide (i.e. X=0, t=0 and x=0).

3. Assume that the speed of something, as measured in both stationary and moving systems is the same and equal to C regardless of the direction of the movement (that "something" is the light in vacuum, but it's physical characteristics are unimportant)

4. Assume further that we are looking for linear orthogonal (i.e. preserving the distance between points and angles between vectors) transformation of coordinates from (X,T) to (x,t) that satisfies the above criteria. What would this transformation be?

Linear transformation from {X,T} system to {x,t} system should look like this:

x = pX + qT

t = rX + sT

where p, q, r and s are 4 unknown coefficients of transformation, which we are going to determine by constructing a system of 4 linear equations with them.

We should not add any constants into above transformations since {X=0,T=0} should transform into {x=0,t=0}.

A. Notice that the property of orthogonality is needed to preserve geometry (i.e. no deformation) and, therefore, to preserve the form of all physical equations of motion. As it is well known, orthogonal transformations have determinant of the matrix of coefficients equal to 1, i.e.

ps − qr = 1. An unfamiliar with this property student can study this subject separately (we at Unizor plan to include this into corresponding topic on vectors).

The above is the first equation to determine unknown coefficients.

B. Since moving system moves along X-axis with speed V, its beginning of coordinate (point x=0) must at the moment of time T be on a distance VT from the beginning of coordinates of a stationary system. Hence, if X=VT, x=0 for any T. From this and the first transformation equation x = pX + qT we derive:

0 = pVT + qT or

0 = (pV + q)T.

Since this equality is true for any T,

pV + q = 0

and, unconditionally, q = −pV.

This is the second equation for our unknown coefficients.

C. Since the speed of light C is the same in both systems {X,T} and {x,t}, an equation of its motion in the stationary system must be X = CT and in the moving system x = Ct. Therefore, if X=CT, then x=Ct. Put X=CT into equations of transformation of coordinates. We get x = pCT + qT, t = rCT + sT. Substitute these expressions into x=Ct:

pCT + qT = r(C^2)T + sCT.

Reduce by T,

pC + q = rC^2 + sC.

This is the third equation for unknown coefficients.

D. Repeat the logic of a previous paragraph for the light moving in the opposite direction with a speed −C. We get, if X = −CT, then x = −Ct. Therefore, x = −pCT + qT, t = −rCT + sT and (since x = −Ct)

−pCT + qT = r(C^2)T − sCT.

Reduce by T,

−pC + q = rC^2 − sC.

This the fourth equation for unknown coefficients.

So, this is the system of 4 linear equations for 4 unknown coefficients of transformation p, q, r, s:

(a) ps − qr = 1

(b) q = −pV

(c) pC + q = rC^2 + sC

(d) −pC + q = rC^2 − sC

Equations (b), (c) and (d) are linear. We can solve them for q, r and s in terms of one variable p. Then we can substitute it into the equation (a) getting a quadratic equation for p and then find solutions for q, r and s.

Variable q is already represented in terms of p in equation (b): q = -pV.

From (c) and (d), adding and subtracting these equations, we get:

2q = 2rC^2, therefore q = rC^2 and r = −pV/C^2

2pC = 2sC, therefore s = p.

Substituting q, r and s, expressed in terms of p, into an equation (a) ps − qr = 1, we get:

p2 − (−pV)(−pV)/C^2 = 1, therefore

p2·(1 − V^2/C^2) = 1 and

p = 1/√(1−(V^2)/(C^2)).

From this all other coefficients of a transformation matrix are derived:

q = −V/√(1−(V^2)/(C^2))

r = −(V/(C^2))/√(1−(V^2)/(C^2))

s = 1/√(1−(V^2)/(C^2))

The final form of transformation of coordinates in the Special Theory of Relativity is:

x = (X − VT)/√(1−(V/C)^2)

t = (T − VX/(C^2))/√(1−(V/C)^2)

1. Assume we have two systems of coordinates, one stationary with coordinates {X,T} (assuming for simplicity all the movements will occur in one space dimension along X-axis and one time dimension) and another with coordinates {x,t} moving along the X-axis with constant speed V.

2. Assume that at time T=0 systems coincide (i.e. X=0, t=0 and x=0).

3. Assume that the speed of something, as measured in both stationary and moving systems is the same and equal to C regardless of the direction of the movement (that "something" is the light in vacuum, but it's physical characteristics are unimportant)

4. Assume further that we are looking for linear orthogonal (i.e. preserving the distance between points and angles between vectors) transformation of coordinates from (X,T) to (x,t) that satisfies the above criteria. What would this transformation be?

Linear transformation from {X,T} system to {x,t} system should look like this:

x = pX + qT

t = rX + sT

where p, q, r and s are 4 unknown coefficients of transformation, which we are going to determine by constructing a system of 4 linear equations with them.

We should not add any constants into above transformations since {X=0,T=0} should transform into {x=0,t=0}.

A. Notice that the property of orthogonality is needed to preserve geometry (i.e. no deformation) and, therefore, to preserve the form of all physical equations of motion. As it is well known, orthogonal transformations have determinant of the matrix of coefficients equal to 1, i.e.

ps − qr = 1. An unfamiliar with this property student can study this subject separately (we at Unizor plan to include this into corresponding topic on vectors).

The above is the first equation to determine unknown coefficients.

B. Since moving system moves along X-axis with speed V, its beginning of coordinate (point x=0) must at the moment of time T be on a distance VT from the beginning of coordinates of a stationary system. Hence, if X=VT, x=0 for any T. From this and the first transformation equation x = pX + qT we derive:

0 = pVT + qT or

0 = (pV + q)T.

Since this equality is true for any T,

pV + q = 0

and, unconditionally, q = −pV.

This is the second equation for our unknown coefficients.

C. Since the speed of light C is the same in both systems {X,T} and {x,t}, an equation of its motion in the stationary system must be X = CT and in the moving system x = Ct. Therefore, if X=CT, then x=Ct. Put X=CT into equations of transformation of coordinates. We get x = pCT + qT, t = rCT + sT. Substitute these expressions into x=Ct:

pCT + qT = r(C^2)T + sCT.

Reduce by T,

pC + q = rC^2 + sC.

This is the third equation for unknown coefficients.

D. Repeat the logic of a previous paragraph for the light moving in the opposite direction with a speed −C. We get, if X = −CT, then x = −Ct. Therefore, x = −pCT + qT, t = −rCT + sT and (since x = −Ct)

−pCT + qT = r(C^2)T − sCT.

Reduce by T,

−pC + q = rC^2 − sC.

This the fourth equation for unknown coefficients.

So, this is the system of 4 linear equations for 4 unknown coefficients of transformation p, q, r, s:

(a) ps − qr = 1

(b) q = −pV

(c) pC + q = rC^2 + sC

(d) −pC + q = rC^2 − sC

Equations (b), (c) and (d) are linear. We can solve them for q, r and s in terms of one variable p. Then we can substitute it into the equation (a) getting a quadratic equation for p and then find solutions for q, r and s.

Variable q is already represented in terms of p in equation (b): q = -pV.

From (c) and (d), adding and subtracting these equations, we get:

2q = 2rC^2, therefore q = rC^2 and r = −pV/C^2

2pC = 2sC, therefore s = p.

Substituting q, r and s, expressed in terms of p, into an equation (a) ps − qr = 1, we get:

p2 − (−pV)(−pV)/C^2 = 1, therefore

p2·(1 − V^2/C^2) = 1 and

p = 1/√(1−(V^2)/(C^2)).

From this all other coefficients of a transformation matrix are derived:

q = −V/√(1−(V^2)/(C^2))

r = −(V/(C^2))/√(1−(V^2)/(C^2))

s = 1/√(1−(V^2)/(C^2))

The final form of transformation of coordinates in the Special Theory of Relativity is:

x = (X − VT)/√(1−(V/C)^2)

t = (T − VX/(C^2))/√(1−(V/C)^2)

Zor Shekhtman

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